Question

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A copper cube is cooled from an initial temperature of 375 °C to a final temperature of 105 °C. By what percentage does its side length change? By what percentage does its volume change? By what percentage does its density change? For copper, the coefficient of linear expansion is 17.0 x 10-6 °C-1 copper cube

Answer 1

Intral Temperature = 375°C final Tempestre - 105°C - Coefficient of linear Exponsion = 17x10 6 oct By what percentage does its side length change. Change in Length (AL) & LOAT = 2LOAT I ALE & LOAT AL = XAT (17X10"")*(375-105) 1 = (17X164) x(375-1051) = 4.59810-3 A = 4.598103 % Change in length = 4.59815 3 xloo o 0.45910

we have AL= LAAT L-Lo= LOCAT Le to + LOKAT LO(1+XAT) L - Lo ( 1+ (17x10=(x (37540s))) L'e Loſ 1+ ( 17810*X(375–105)) lol + 0.004597 L' - Lo(1+4:59 x103) How volume of cube having side length Lo Vo - Los

Now volume of cube having side length to Vo = Lo² After temperature ch v (1764.59800)) 1.3 Vo=L03 VIP (1+ (4 saxls")) Los = (1.014) Lo3 = (2014 Dvo AVV'- Vo = 1.014 L² - L3 = Lo3 (1.014-1) ololu vo Ava to 014) Vo AV 0.014 2 1.4°°||

Density = mass Volume M (lolm) Vo 1.014 AS = fl-lo = So - so 1014 1014 T TAS = fol11 -1 11.014 Aste fo torotu) We are only studying the magnitude the of change 1014 0.014 1.014 1471 104 som to 1981x100 = 1:38

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